CBSE Class 12 Physics Question Paper 2025 Solved Code 55-1-2

SECTION A

1. In the figure curved lines represent equipotential surfaces. A charge Q is moved along different paths A, B, C and D. The work done on the charge will be maximum along the path [1 Mark]
(A) A
(B) B
(C) C
(D) D

[Figure: Four curved equipotential surfaces labelled 10 V, 15 V, 20 V and 25 V. The charge Q lies on the 10 V surface. Path A goes from Q to another point on the 10 V surface, path B ends on the 15 V surface, path C ends on the 25 V surface and path D ends on the 20 V surface.]

Answer: (C) C

Teacher's Note:
a) Work done in moving a charge is \( W = q\Delta V \), so it depends only on the potential difference between the start and end points.
b) Path C takes the charge from 10 V to 25 V, the largest change of potential (15 V), so the work done is maximum along C.

2. The resistance of a wire of length L and radius r is R. Which one of the following would provide a wire of the same material of resistance \( \frac{R}{2} \) ? [1 Mark]
(A) Using a wire of same radius and twice the length
(B) Using a wire of same radius and half length
(C) Using a wire of same length and twice the radius
(D) Using a wire of same length and half the radius

Answer: (B) Using a wire of same radius and half length

Teacher's Note:
a) Resistance \( R = \frac{\rho L}{\pi r^2} \), so R is directly proportional to length for the same radius.
b) Halving the length halves the resistance; doubling the radius would make it \( \frac{R}{4} \), not \( \frac{R}{2} \).

3. A 1 cm segment of a wire lying along x-axis carries current of 0.5 A along +x direction. A magnetic field \( \vec{B} = (0.4\text{ mT})\hat{j} + (0.6\text{ mT})\hat{k} \) is switched on, in the region. The force acting on the segment is [1 Mark]
(A) \( (2\hat{j} + 3\hat{k}) \) mN
(B) \( (-3\hat{j} + 2\hat{k}) \) μN
(C) \( (6\hat{j} + 4\hat{k}) \) mN
(D) \( (-4\hat{j} + 6\hat{k}) \) μN

Answer: (B) \( (-3\hat{j} + 2\hat{k}) \) μN

Teacher's Note:
a) Use \( \vec{F} = I(\vec{L} \times \vec{B}) \) with \( \vec{L} = 0.01\hat{i} \) m.
b) Remember \( \hat{i} \times \hat{j} = \hat{k} \) and \( \hat{i} \times \hat{k} = -\hat{j} \) to get the correct signs.

4. A circular coil of diameter 15 mm having 300 turns is placed in a magnetic field of 30 mT such that the plane of the coil is perpendicular to the direction of magnetic field. The magnetic field is reduced uniformly to zero in 20 ms and again increased uniformly to 30 mT in 40 ms. If the emfs induced in the two time intervals are e₁ and e₂ respectively, then the value of e₁/e₂ is [1 Mark]
(A) \( \frac{1}{2} \)
(B) 1
(C) 2
(D) 4

Answer: (C) 2

Teacher's Note:
a) Induced emf \( e = \frac{N\Delta\phi}{\Delta t} \), and the change in flux is the same in both intervals.
b) So \( \frac{e_1}{e_2} = \frac{\Delta t_2}{\Delta t_1} = \frac{40}{20} = 2 \); the diameter and number of turns are not needed.

5. You are required to design an air-filled solenoid of inductance 0.016 H having a length 0.81 m and radius 0.02 m. The number of turns in the solenoid should be [1 Mark]
(A) 2592
(B) 2866
(C) 2976
(D) 3140

Answer: (B) 2866

Teacher's Note:
a) Use \( L = \frac{\mu_0 N^2 A}{l} \), so \( N^2 = \frac{Ll}{\mu_0 A} \) with \( A = \pi r^2 \).
b) Substituting the values gives \( N \approx 2866 \) turns.

6. A voltage \( v = v_0 \sin \omega t \) applied to a circuit drives a current \( i = i_0 \sin(\omega t + \phi) \) in the circuit. The average power consumed in the circuit over a cycle is [1 Mark]
(A) Zero
(B) \( i_0 v_0 \cos\phi \)
(C) \( \frac{i_0 v_0}{2} \)
(D) \( \frac{i_0 v_0}{2}\cos\phi \)

Answer: (D) \( \frac{i_0 v_0}{2}\cos\phi \)

Teacher's Note:
a) Average power in an AC circuit is \( P_{avg} = V_{rms}I_{rms}\cos\phi \).
b) Using \( V_{rms} = \frac{v_0}{\sqrt2} \) and \( I_{rms} = \frac{i_0}{\sqrt2} \) gives \( \frac{i_0v_0}{2}\cos\phi \).

7. Which one of the following correctly represents the change in wave characteristics (all in vacuum) from microwaves to X-rays in electromagnetic spectrum ? [1 Mark]
Speed | Wavelength | Frequency
(A) Remains same | Decreases | Remains same
(B) Remains same | Decreases | Increases
(C) Increases | Increases | Decreases
(D) Remains same | Increases | Remains same

Answer: (B) Remains same | Decreases | Increases

Teacher's Note:
a) All electromagnetic waves travel with the same speed c in vacuum.
b) From microwaves to X-rays the wavelength decreases, so by \( c = \nu\lambda \) the frequency increases.

8. The speed of light in two media '1' and '2' are v₁ and v₂ \( (\gt v_1) \) respectively. For a ray of light to undergo total internal reflection at the interface of these two media, it must be incident from [1 Mark]
(A) medium '1' and at an angle greater than \( \sin^{-1}\left(\frac{v_1}{v_2}\right) \)
(B) medium '1' and at an angle greater than \( \cos^{-1}\left(\frac{v_1}{v_2}\right) \)
(C) medium '2' and at an angle greater than \( \sin^{-1}\left(\frac{v_1}{v_2}\right) \)
(D) medium '2' and at an angle greater than \( \cos^{-1}\left(\frac{v_1}{v_2}\right) \)

Answer: (A) medium '1' and at an angle greater than \( \sin^{-1}\left(\frac{v_1}{v_2}\right) \)

Teacher's Note:
a) Light is slower in medium 1, so medium 1 is optically denser; total internal reflection needs the ray to go from the denser to the rarer medium.
b) The critical angle is given by \( \sin C = \frac{n_2}{n_1} = \frac{v_1}{v_2} \), and the angle of incidence must be greater than C.

9. A source produces monochromatic light of frequency \( 5.0 \times 10^{14} \) Hz and the power emitted is 3.31 mW. The number of photons emitted per second by the source, on an average is [1 Mark]
(A) 10¹⁶
(B) 10²⁴
(C) 10¹⁰
(D) 10²⁰

Answer: (A) 10¹⁶

Teacher's Note:
a) Number of photons per second \( n = \frac{P}{h\nu} \).
b) \( n = \frac{3.31 \times 10^{-3}}{6.63 \times 10^{-34} \times 5 \times 10^{14}} \approx 10^{16} \) per second.

10. Which of the following figures correctly represent the shape of curve of binding energy per nucleon as a function of mass number ? [1 Mark]

[Figure: Four graphs of B.E./A (y-axis) versus mass number A (x-axis). (A) rises steeply, reaches a broad peak near A = 56 and then falls gradually. (B) rises steeply and then stays almost flat after A = 56. (C) is a symmetric hump peaking at A = 80. (D) rises sharply to a peak at A = 80, then falls sharply and rises slightly.]

Answer: (A) the curve that rises sharply, peaks near A = 56 and then decreases gradually

Teacher's Note:
a) The binding energy per nucleon curve has its maximum (about 8.8 MeV) near A = 56, the iron region.
b) For heavier nuclei it falls off slowly, so a flat curve or a peak at A = 80 is wrong.

11. When a p-n junction diode is forward biased [1 Mark]
(A) the barrier height and the depletion layer width both increase.
(B) the barrier height increases and the depletion layer width decreases.
(C) the barrier height and the depletion layer width both decrease.
(D) the barrier height decreases and the depletion layer width increases.

Answer: (C) the barrier height and the depletion layer width both decrease.

Teacher's Note:
a) In forward bias the applied voltage opposes the barrier potential, so the barrier height decreases.
b) A lower barrier needs fewer uncovered ions, so the depletion layer also becomes thinner.

12. Let \( \lambda_e \), \( \lambda_p \) and \( \lambda_d \) be the wavelengths associated with an electron, a proton and a deuteron, all moving with the same speed. Then the correct relation between them is [1 Mark]
(A) \( \lambda_d \gt \lambda_p \gt \lambda_e \)
(B) \( \lambda_e \gt \lambda_p \gt \lambda_d \)
(C) \( \lambda_p \gt \lambda_e \gt \lambda_d \)
(D) \( \lambda_e = \lambda_p = \lambda_d \)

Answer: (B) \( \lambda_e \gt \lambda_p \gt \lambda_d \)

Teacher's Note:
a) de Broglie wavelength \( \lambda = \frac{h}{mv} \), so for the same speed \( \lambda \propto \frac{1}{m} \).
b) Mass increases from electron to proton to deuteron, so the wavelength decreases in the same order.

Note : Question numbers 13 to 16 are Assertion (A) and Reason (R) type questions. Two statements are given - one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the codes (A), (B), (C) and (D) as given below.

13. Assertion (A) : The potential energy of an electron revolving in any stationary orbit in a hydrogen atom is positive.
Reason (R) : The total energy of a charged particle is always positive. [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false and Reason (R) is also false.

Answer: (D) Assertion (A) is false and Reason (R) is also false.

Teacher's Note:
a) The electron's potential energy in a hydrogen atom is negative because the electron-nucleus force is attractive.
b) The total energy of a charged particle can be positive, negative or zero; for a bound electron it is negative.

14. Assertion (A) : We cannot form a p-n junction diode by taking a slab of a p-type semiconductor and physically joining it to another slab of a n-type semiconductor.
Reason (R) : In a p-type semiconductor \( \eta_e \gt\gt \eta_h \) while in a n-type semiconductor \( \eta_h \gt\gt \eta_e \). [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false and Reason (R) is also false.

Answer: (C) Assertion (A) is true, but Reason (R) is false.

Teacher's Note:
a) Physically joining two slabs leaves a rough, discontinuous surface, so a proper junction is not formed; the Assertion is true.
b) The Reason is false because holes are the majority carriers in p-type and electrons in n-type semiconductors.

15. Assertion (A) : The deflection in a galvanometer is directly proportional to the current passing through it.
Reason (R) : The coil of a galvanometer is suspended in a uniform radial magnetic field. [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false and Reason (R) is also false.

Answer: (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

Teacher's Note:
a) In a radial field the plane of the coil is always parallel to B, so the torque NIAB is directly proportional to I.
b) This makes the deflection proportional to the current and gives a linear scale.

16. Assertion (A) : It is difficult to move a magnet into a coil of large number of turns when the circuit of the coil is closed.
Reason (R) : The direction of induced current in a coil with its circuit closed, due to motion of a magnet, is such that it opposes the cause. [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false and Reason (R) is also false.

Answer: (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

Teacher's Note:
a) This is Lenz's law: the induced current opposes the motion of the magnet.
b) More turns give a larger induced current and a stronger opposing force, so the magnet is harder to push in.

SECTION B

17. Show that \( \vec{E} = \rho\vec{j} \) leads to Ohm's law. Write a condition in which the Ohm's law is not valid for a material. [2 Marks]

Answer:
1. For a conductor of length l and area A, \( E = \frac{V}{l} \) and \( j = \frac{I}{A} \). So \( \vec{E} = \rho\vec{j} \) gives \( \frac{V}{l} = \rho\frac{I}{A} \).
2. Hence \( \frac{V}{I} = \frac{\rho l}{A} \) = constant (R), that is V = IR, which is Ohm's law.
3. Ohm's law is not valid at high temperature, or for a semiconductor.

Teacher's Note:
a) Write E = V/l and j = I/A clearly; this substitution carries the marks.
b) State that \( \frac{\rho l}{A} \) is constant for a given conductor, which is the resistance R.
c) Only one condition is asked; high temperature or semiconductors are accepted.

18. (a) In a diffraction experiment, the slit is illuminated by light of wavelength 600 nm. The first minimum of the pattern falls at \( \theta = 30^{\circ} \). Calculate the width of the slit. [2 Marks]

Answer:
1. Condition for minima: \( a\sin\theta = n\lambda \). For the first minimum, n = 1.
2. \( a \sin 30^{\circ} = 600 \times 10^{-9} \) m, so \( a \times \frac{1}{2} = 600 \times 10^{-9} \) m.
3. \( a = 1200 \times 10^{-9} \) m \( = 1.2 \times 10^{-6} \) m.

Teacher's Note:
a) Use the single-slit minima condition \( a\sin\theta = n\lambda \), not the interference condition.
b) Use \( \sin 30^{\circ} = \frac{1}{2} \) and convert nm to m.

OR

(b) In a Young's double-slit experiment, two light waves, each of intensity \( I_0 \), interfere at a point, having a path difference \( \frac{\lambda}{8} \) on the screen. Find the intensity at this point. [2 Marks]

Answer:
1. Phase difference \( \Delta\phi = \frac{2\pi}{\lambda}\Delta x = \frac{2\pi}{\lambda}\times\frac{\lambda}{8} = \frac{\pi}{4} \).
2. \( I = I_0 + I_0 + 2\sqrt{I_0I_0}\cos\frac{\pi}{4} = 2I_0 + 2I_0 \times \frac{1}{\sqrt2} \).
3. \( I = I_0(2+\sqrt2) = 3.414\, I_0 \) (or \( I = 4I_0\cos^2\frac{\pi}{8} \)).

Teacher's Note:
a) First convert path difference into phase difference using \( \frac{2\pi}{\lambda} \).
b) Either \( I = I_1 + I_2 + 2\sqrt{I_1I_2}\cos\phi \) or \( I = 4I_0\cos^2\frac{\phi}{2} \) is accepted.

19. A spherical convex surface of radius of curvature R separates glass (refractive index 1.5) from air. Light from a point source placed in air at distance R/2 from the surface falls on it. Find the position and nature of the image formed. [2 Marks]

Answer:
1. Refraction is from rarer to denser medium: \( \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} \), with \( u = -\frac{R}{2} \), \( n_1 = 1 \), \( n_2 = 1.5 \).
2. \( \frac{1.5}{v} + \frac{2}{R} = \frac{0.5}{R} \), so \( \frac{1.5}{v} = -\frac{1.5}{R} \), giving v = -R.
3. The image is virtual, formed in air (on the same side as the object) at a distance R from the surface.

Teacher's Note:
a) Take u negative for a real object in air and R positive for a surface convex towards the object.
b) A negative v means the image is on the object's side, so it is virtual.

20. The energy of an electron in an orbit of Bohr hydrogen atom is -3.4 eV. Find its angular momentum. [2 Marks]

Answer:
1. \( E_n = -\frac{13.6}{n^2} \) eV, so \( n^2 = \frac{-13.6}{-3.4} = 4 \), giving n = 2.
2. Angular momentum \( L = \frac{nh}{2\pi} = \frac{2h}{2\pi} = \frac{h}{\pi} \).
3. \( L = \frac{6.63\times10^{-34}}{3.14} = 2.11 \times 10^{-34} \) Js.

Teacher's Note:
a) First find the orbit number n from the energy, then use Bohr's quantisation condition.
b) Write the unit Js (or kg m² s^-1) with the final answer.

21. A p-type Si semiconductor is made by doping an average of one dopant atom per \( 5 \times 10^7 \) silicon atoms. If the number density of silicon atoms in the specimen is \( 5 \times 10^{28} \) atoms m^-3, find the number of holes created per cubic centimetre in the specimen due to doping. Also give one example of such dopants. [2 Marks]

Answer:
1. Number of holes created per m³ \( = \frac{5\times10^{28}}{5\times10^7} = 10^{21} \), so per cm³ it is \( \frac{10^{21}}{10^6} = 10^{15} \).
2. One example of such a dopant: aluminium (or indium or gallium).

Teacher's Note:
a) Each trivalent dopant atom creates one hole, so divide the Si density by the doping ratio.
b) Remember \( 1 \text{ m}^3 = 10^6 \text{ cm}^3 \) while converting.

SECTION C

22. (a) Two batteries of emf's 3V & 6V and internal resistances 0.2 Ω & 0.4 Ω are connected in parallel. This combination is connected to a 4 Ω resistor. Find :
(i) the equivalent emf of the combination
(ii) the equivalent internal resistance of the combination
(iii) the current drawn from the combination [3 Marks]

Answer:
1. Equivalent emf \( E_{eq} = \frac{E_1r_2+E_2r_1}{r_1+r_2} = \frac{3\times0.4+6\times0.2}{0.6} = 4 \) V.
2. Equivalent internal resistance \( r_{eq} = \frac{r_1r_2}{r_1+r_2} = \frac{0.2\times0.4}{0.2+0.4} = 0.133 \) Ω.
3. Current drawn \( I = \frac{E_{eq}}{R+r_{eq}} = \frac{4}{4+0.133} = \frac{4}{4.133} \approx 0.97 \) A.

Teacher's Note:
a) Learn the formulas for two cells in parallel: \( E_{eq} = \frac{E_1r_2+E_2r_1}{r_1+r_2} \) and \( r_{eq} = \frac{r_1r_2}{r_1+r_2} \).
b) Then treat the combination as one cell of emf \( E_{eq} \) and internal resistance \( r_{eq} \) in series with the 4 Ω resistor.
c) The marking scheme writes the final current as 0.9 A; the exact value of 4/4.133 is about 0.97 A.

OR

(b) (i) A conductor of length \( l \) is connected across an ideal cell of emf E. Keeping the cell connected, the length of the conductor is increased to \( 2l \) by gradually stretching it. If R and R' are initial and final values of resistance and v_d and v_d' are initial and final values of drift velocity, find the relation between (i) R' and R and (ii) v_d' and v_d.
(ii) When electrons drift in a conductor from lower to higher potential, does it mean that all the 'free electrons' of the conductor are moving in the same direction ? [3 Marks]

Answer:
1. Volume stays constant: \( Al = A'(2l) \), so \( A' = \frac{A}{2} \). Then \( R' = \frac{\rho(2l)}{A/2} = \frac{4\rho l}{A} \), so \( \frac{R'}{R} = 4 \), that is R' = 4R.
2. \( v_d = \frac{eE}{m}\tau = \frac{eV}{ml}\tau \) and \( v_d' = \frac{eV}{ml'}\tau \), so \( \frac{v_d'}{v_d} = \frac{l}{l'} = \frac{1}{2} \), that is \( v_d' = \frac{v_d}{2} \).
3. No. Drift is only a small average velocity; the free electrons still move randomly in all directions.

Teacher's Note:
a) On stretching, the volume of the wire is constant, so doubling the length halves the area.
b) The cell (V) stays the same, so the electric field \( E = \frac{V}{l} \) halves when the length doubles.
c) A short answer "No" with a reason about random motion earns the mark in part (ii).

23. (a) Define magnetic moment of a current-carrying coil. Write its SI unit.
(b) A coil of 60 turns and area \( 1.5 \times 10^{-3} \) m³ carrying 2A current lies in a vertical plane. It experiences a torque of 0.12 Nm when placed in a uniform horizontal magnetic field. The torque acting on the coil changes to 0.05 Nm after the coil is rotated about its diameter by \( 90^{\circ} \), in the magnetic field. Find the magnitude of the magnetic field. [3 Marks]

Answer:
1. Magnetic moment of a current-carrying coil is the product of the current flowing through the coil and the area of the coil \( (\vec{M} = I\vec{A}) \). Its SI unit is A m².
2. \( \tau = NIAB\sin\theta \). First position: \( 0.12 = 60\times2\times1.5\times10^{-3}\times B\sin\theta \), so \( B\sin\theta = \frac{2}{3} \).
3. After rotation by \( 90^{\circ} \): \( 0.05 = 60\times2\times1.5\times10^{-3}\times B\cos\theta \), so \( B\cos\theta = \frac{5}{18} \).
4. \( B = \sqrt{B^2\sin^2\theta + B^2\cos^2\theta} = \sqrt{\left(\frac{2}{3}\right)^2 + \left(\frac{5}{18}\right)^2} = \frac{13}{18} \) T \( \approx 0.72 \) T.

Teacher's Note:
a) Rotating the coil by \( 90^{\circ} \) changes \( \sin\theta \) into \( \cos\theta \); squaring and adding removes \( \theta \).
b) NIA = 0.18 here, so divide each torque by 0.18 to get \( B\sin\theta \) and \( B\cos\theta \).
c) The paper prints the area unit as m³; it should be read as m².

24. Consider two long co-axial solenoids S₁ and S₂, each of length \( l \) \( (\gt\gt r_2) \) and of radius r₁ and r₂ \( (r_2 \gt r_1) \). The number of turns per unit length are n₁ and n₂ respectively. Derive an expression for mutual inductance M₁₂ of solenoid S₁ with respect to solenoid S₂. Show that M₂₁ = M₁₂. [3 Marks]

Answer:
1. Diagram: inner solenoid S₁ (radius r₁, N₁ turns) placed coaxially inside outer solenoid S₂ (radius r₂, N₂ turns), both of length l.
2. When current I₂ flows in S₂, the field inside is \( \mu_0 n_2 I_2 \). Flux linkage with S₁: \( N_1\phi_1 = M_{12}I_2 \) and \( N_1\phi_1 = (n_1 l)(\pi r_1^2)(\mu_0 n_2 I_2) \).
3. Comparing, \( M_{12} = \mu_0 n_1 n_2 \pi r_1^2 l \).
4. Reverse case: current I₁ in S₁, field \( \mu_0 n_1 I_1 \) exists only inside S₁ (area \( \pi r_1^2 \)). Flux linkage with S₂: \( N_2\phi_2 = M_{21}I_1 = (n_2 l)(\pi r_1^2)(\mu_0 n_1 I_1) \).
5. So \( M_{21} = \mu_0 n_1 n_2 \pi r_1^2 l \), and therefore \( M_{12} = M_{21} \).

Teacher's Note:
a) In both cases use the area of the inner solenoid \( (\pi r_1^2) \), because the field of S₁ exists only inside it.
b) Write the two flux-linkage equations separately and compare them; this comparison carries the last mark.
c) The marking scheme's last line for M₂₁ leaves out \( \mu_0 \); the correct expression includes it.

25. (a) A parallel plate capacitor is charged by an ac source. Show that the sum of conduction current (I_c) and the displacement current (I_d) has the same value at all points of the circuit.
(b) In case (a) above, is Kirchhoff's first rule (junction rule) valid at each plate of the capacitor ? Explain. [3 Marks]

Answer:
1. Total current \( I = I_c + I_d \). Outside the capacitor, \( I_d = 0 \), so \( I = I_c \).
2. Inside the capacitor, \( I_c = 0 \), so \( I = I_d = \varepsilon_0\frac{d\phi_E}{dt} = \varepsilon_0\frac{d}{dt}(EA) = \varepsilon_0\frac{d}{dt}\left(\frac{\sigma}{\varepsilon_0}A\right) = A\frac{d}{dt}\left(\frac{Q}{A}\right) \).
3. So \( I = \frac{dQ}{dt} = I_c \); hence \( I_c + I_d \) has the same value at all points of the circuit.
4. (b) Yes. The current entering the plate is I_c and the current between the plates is I_d; since \( I_c = I_d \), Kirchhoff's junction rule is valid at each plate.

Teacher's Note:
a) Consider the two regions separately: only conduction current in the wires and only displacement current between the plates.
b) Use \( E = \frac{\sigma}{\varepsilon_0} = \frac{Q}{\varepsilon_0 A} \) to show that \( I_d = \frac{dQ}{dt} \).

26. (a) Draw a plot of frequency ν of incident radiations as a function of stopping potential V₀ for a given photo emissive material. What information can be obtained from the value of the intercept on the stopping potential axis ?
(b) Calculate : (i) the momentum and (ii) de Broglie wavelength, of an electron with kinetic energy of 80 eV. [3 Marks]

Answer:
1. The graph between stopping potential V₀ and frequency ν is a straight line. It cuts the frequency axis at the threshold frequency \( \nu_0 \), and when extended it cuts the stopping potential axis at \( -\frac{\phi_0}{e} \).
2. The value of the work function \( \phi_0 \) of the material can be obtained from the intercept on the stopping potential axis.
3. (b)(i) \( p = \sqrt{2mK} = \sqrt{2\times9.1\times10^{-31}\times80\times1.6\times10^{-19}} = 4.8\times10^{-24} \) kg m/s.
4. (b)(ii) \( \lambda = \frac{h}{p} = \frac{6.63\times10^{-34}}{4.8\times10^{-24}} = 1.38\times10^{-10} \) m.

Teacher's Note:
a) The straight-line graph follows from \( eV_0 = h\nu - \phi_0 \); show the extended (dotted) part meeting the V₀ axis.
b) Convert 80 eV into joules \( (80\times1.6\times10^{-19}) \) before using \( p = \sqrt{2mK} \).
c) The marking scheme writes \( \lambda = 1.38\times10^{-9} \) m (using \( 4.8\times10^{-25} \)); with \( p = 4.8\times10^{-24} \) kg m/s the correct value is \( 1.38\times10^{-10} \) m.

27. (a) Draw circuit arrangement for studying V-I characteristics of a p-n junction diode.
(b) Show the shape of the characteristics of a diode.
(c) Mention two information that you can get from these characteristics. [3 Marks]

Answer:
1. Forward bias circuit: the p-side of the diode is joined to the positive terminal of a variable battery through a milliammeter and a switch, with a voltmeter across the diode. Reverse bias circuit: the diode connections are reversed and a microammeter is used.
2. Characteristic curve: in forward bias the current stays very small until the knee (threshold) voltage and then rises sharply (in mA); in reverse bias a very small, almost constant current (in μA) flows until the breakdown voltage, where it increases suddenly.
3. Two informations: the knee voltage and the reverse saturation current (the breakdown voltage, very low resistance in forward bias or very high resistance in reverse bias are also accepted).

Teacher's Note:
a) Use a milliammeter for forward bias and a microammeter for reverse bias, because the currents differ greatly.
b) As per the marking scheme, marks are not deducted for not writing values on the graph.

28. (a) Define 'Mass defect' and 'Binding energy' of a nucleus. Describe 'Fission process' on the basis of binding energy per nucleon.
(b) A deuteron contains a proton and a neutron and has a mass of 2.013553 u. Calculate the mass defect for it in u and its energy equivalence in MeV. \( (m_p = 1.007277 \text{ u}, m_n = 1.008665 \text{ u}, 1u = 931.5 \text{ MeV/c}^2) \) [3 Marks]

Answer:
1. Mass defect is the difference between the total mass of the constituents (nucleons) and the mass of the nucleus.
2. Binding energy is the energy required to separate the nucleons from the nucleus.
3. In fission, a heavy nucleus splits into lighter nuclei and energy is released; as a result the binding energy per nucleon increases.
4. \( \Delta m = (m_p + m_n) - m_d = (1.007277 + 1.008665) - 2.013553 = 0.002389 \) u.
5. Energy released \( = \Delta m \times c^2 = 0.002389 \times 931.5 = 2.2253 \) MeV \( \approx 2.22 \) MeV.

Teacher's Note:
a) Link fission to the binding energy curve: the products have higher binding energy per nucleon than the heavy parent nucleus.
b) Keep all six decimal places while subtracting masses, otherwise the small mass defect goes wrong.

SECTION D

Question numbers 29 and 30 are case study based questions. Read the following paragraphs and answer the questions that follow.

29. A thin lens is a transparent optical medium bounded by two surfaces, at least one of which should be spherical. Applying the formula for image formation by a single spherical surface successively at the two surfaces of a lens, one can obtain the 'lens maker formula' and then the 'lens formula'. A lens has two foci - called 'first focal point' and 'second focal point' of the lens, one on each side. [4 Marks]

(i) Consider the arrangement shown in figure. A black vertical arrow and a horizontal thick line with a ball are painted on a glass plate. It serves as the object. When the plate is illuminated, its real image is formed on the screen.
Which of the following correctly represents the image formed on the screen ? [1 Mark]

[Figure: An optical bench with a screen at one end, a convex lens in the middle and a light box with a plate at the other end. The plate, on the hidden side of the box, shows a vertical arrow pointing up and a horizontal line with a ball at its right end. Options: (A) arrow pointing up with the ball on the right; (B) arrow pointing up with a slanting line and the ball at its lower left; (C) arrow pointing down with the ball on the left; (D) arrow pointing down with a slanting line and the ball at its upper right.]

Answer: (C) the image with the arrow pointing down and the ball on the left

Teacher's Note:
a) A real image formed by a convex lens is inverted, so the arrow points down.
b) The image is also laterally inverted, so the ball shifts from the right side to the left side.

(ii) Which of the following statements is incorrect ? [1 Mark]
(A) For a convex mirror magnification is always negative.
(B) For all virtual images formed by a mirror magnification is positive.
(C) For a concave lens magnification is always positive.
(D) For real and inverted images, magnification is always negative.

Answer: (A) For a convex mirror magnification is always negative.

Teacher's Note:
a) A convex mirror always forms a virtual, erect image, so its magnification is always positive.
b) Remember: positive magnification means erect (virtual) image, negative means inverted (real) image.

(iii) A convex lens of focal length 'f' is cut into two equal parts perpendicular to the principal axis. The focal length of each part will be : [1 Mark]
(A) f
(B) 2 f
(C) \( \frac{f}{2} \)
(D) \( \frac{f}{4} \)

Answer: (B) 2 f

Teacher's Note:
a) Each part is a plano-convex lens with one curved surface, so by the lens maker formula its power is half.
b) Half the power means double the focal length, that is 2f.

OR

(iii) If an object in case (i) above is 20 cm from the lens and the screen is 50 cm away from the object, the focal length of the lens used is [1 Mark]
(A) 10 cm
(B) 12 cm
(C) 16 cm
(D) 20 cm

Answer: (B) 12 cm

Teacher's Note:
a) Here u = -20 cm and v = 50 - 20 = +30 cm.
b) \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{30} + \frac{1}{20} = \frac{1}{12} \), so f = 12 cm.

(iv) The distance of an object from first focal point of a biconvex lens is X₁ and distance of the image from second focal point is X₂. The focal length of the lens is [1 Mark]
(A) \( X_1 X_2 \)
(B) \( \sqrt{X_1+X_2} \)
(C) \( \sqrt{X_1X_2} \)
(D) \( \sqrt{\frac{X_2}{X_1}} \)

Answer: (C) \( \sqrt{X_1X_2} \)

Teacher's Note:
a) This is Newton's formula for a lens: \( X_1X_2 = f^2 \).
b) It follows from the lens formula by putting u = -(f + X₁) and v = f + X₂.

30. A circuit consisting of a capacitor C, a resistor of resistance R and an ideal battery of emf V, as shown in figure is known as RC series circuit.
As soon as the circuit is completed by closing key S₁ (keeping S₂ open) charges begin to flow between the capacitor plates and the battery terminals. The charge on the capacitor increases and consequently the potential difference V_c (= q/C) across the capacitor also increases with time. When this potential difference equals the potential difference across the battery, the capacitor is fully charged (Q = VC). During this process of charging, the charge q on the capacitor changes with time t as \( q = Q[1 - e^{-t/RC}] \)
The charging current can be obtained by differentiating it and using \( \frac{d}{dx}(e^{mx}) = me^{mx} \).
Consider the case when R = 20 kΩ, C = 500 μF and V = 10 V. [4 Marks]

[Figure: A rectangular circuit with capacitor C in the top arm and resistor R in the left arm. Below R, key S₂ connects directly across to the right arm (discharge path), and key S₁ connects the battery V in the bottom arm (charging path).]

(i) The final charge on the capacitor, when key S₁ is closed and S₂ is open, is [1 Mark]
(A) 5 μC
(B) 5 mC
(C) 25 mC
(D) 0.1 C

Answer: (B) 5 mC

Teacher's Note:
a) Final charge \( Q = VC = 10 \times 500\times10^{-6} = 5\times10^{-3} \) C = 5 mC.
b) Convert μF to F before multiplying.

(ii) For sufficient time the key S₁ is closed and S₂ is open. Now key S₂ is closed and S₁ is open. What is the final charge on the capacitor ? [1 Mark]
(A) Zero
(B) 5 mC
(C) 2.5 mC
(D) 5 μC

Answer: (A) Zero

Teacher's Note:
a) With S₁ open the battery is cut off, and the capacitor discharges through R via S₂.
b) After a long time the discharge is complete, so the final charge is zero.

(iii) The dimensional formula for RC is [1 Mark]
(A) \( [M L^2 T^{-3} A^{-2}] \)
(B) \( [M^0 L^0 T^{-1} A^0] \)
(C) \( [M^{-1} L^{-2} T^4 A^2] \)
(D) \( [M^0 L^0 T A^0] \)

Answer: (D) \( [M^0 L^0 T A^0] \)

Teacher's Note:
a) RC is the time constant of the circuit, so it has the dimension of time.
b) Quick check: the power \( -\frac{t}{RC} \) of e must be dimensionless, so RC has the same dimension as t.

(iv) The key S₁ is closed and S₂ is open. The value of current in the resistor after 5 seconds, is [1 Mark]
(A) \( \frac{1}{2\sqrt{e}} \) mA
(B) \( \sqrt{e} \) mA
(C) \( \frac{1}{\sqrt{e}} \) mA
(D) \( \frac{1}{2e} \) mA

Answer: (A) \( \frac{1}{2\sqrt{e}} \) mA

Teacher's Note:
a) Charging current \( i = \frac{dq}{dt} = \frac{V}{R}e^{-t/RC} \), with \( \frac{V}{R} = 0.5 \) mA and RC = 10 s.
b) At t = 5 s, \( i = 0.5\,e^{-0.5} = \frac{1}{2\sqrt{e}} \) mA.
c) As per the marking scheme, 1 mark for this part may be given to all students who have attempted other parts of the question.

OR

(iv) The key S₁ is closed and S₂ is open. The initial value of charging current in the resistor, is [1 Mark]
(A) 5 mA
(B) 0.5 mA
(C) 2 mA
(D) 1 mA

Answer: (B) 0.5 mA

Teacher's Note:
a) At t = 0 the capacitor is uncharged, so the whole emf acts across R and \( i_0 = \frac{V}{R} \).
b) \( i_0 = \frac{10}{20\times10^3} = 0.5\times10^{-3} \) A = 0.5 mA.

SECTION E

31. (a) (i) (1) What are coherent sources ? Why are they necessary for observing a sustained interference pattern ?
(2) Lights from two independent sources are not coherent. Explain.
(ii) Two slits 0.1 mm apart are arranged 1.20 m from a screen. Light of wavelength 600 nm from a distant source is incident on the slits.
(1) How far apart will adjacent bright interference fringes be on the screen ?
(2) Find the angular width (in degree) of the first bright fringe. [5 Marks]

Answer:
1. Two sources are coherent if the phase difference between the waves from them does not change with time (they emit light of the same frequency with zero or constant phase difference).
2. Coherent sources are needed because only then the phase difference at each point stays constant, so the positions of bright and dark fringes stay fixed and the pattern is sustained.
3. Two independent sources are never coherent because the phase difference between the light emitted by them does not remain constant; it changes randomly with time.
4. Distance between adjacent bright fringes = fringe width \( \beta = \frac{\lambda D}{d} = \frac{600\times10^{-9}\times1.2}{0.1\times10^{-3}} = 7.2 \) mm.
5. Angular width \( \theta = \frac{\lambda}{d} = \frac{600\times10^{-9}}{0.1\times10^{-3}} = 6\times10^{-3} \) rad \( = 0.34^{\circ} \).

Teacher's Note:
a) Use \( \beta = \frac{\lambda D}{d} \) for fringe separation and \( \theta = \frac{\lambda}{d} \) for angular width.
b) To convert radians to degrees, multiply by \( \frac{180}{\pi} \approx 57.3 \).
c) As per the marking scheme, full marks are given even if the angular width is written in radians only.

OR

(b) (i) Define a wavefront. An incident plane wave falls on a convex lens and gets refracted through it. Draw a diagram to show the incident and refracted wavefront.
(ii) A beam of light coming from a distant source is refracted by a spherical glass ball (refractive index 1.5) of radius 15 cm. Draw the ray diagram and obtain the position of the final image formed. [5 Marks]

Answer:
1. A wavefront is the locus of all the points which oscillate in phase.
2. Diagram: plane wavefronts fall on the convex lens from the left; after refraction they emerge as spherical wavefronts of radius f converging to the focus F on the right.
3. Ray diagram: parallel rays enter the glass ball (centre C) at the first surface P₁, bend towards the axis, leave at the second surface P₂ and meet the axis at I beyond P₂.
4. First surface (rarer to denser): n₁ = 1, n₂ = 1.5, R = 15 cm, \( u = \infty \). \( \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} \) gives \( \frac{1.5}{v} - \frac{1}{\infty} = \frac{1.5-1}{15} \), so v = 45 cm from P₁.
5. Second surface (denser to rarer): this image is 45 - 30 = 15 cm beyond P₂, so u = +15 cm and R = -15 cm. \( \frac{1}{v} - \frac{1.5}{15} = \frac{1-1.5}{-15} \) gives v = 7.5 cm. The final image is formed 7.5 cm from the second surface, outside the ball.

Teacher's Note:
a) A wavefront is always perpendicular to the rays at every point.
b) Apply the single-surface refraction formula twice; the image of the first surface is the object for the second surface.
c) Measure the second object distance from P₂ (diameter 30 cm) and take R negative for the second surface.

32. (a) (i) Two point charges 5 μC and -1 μC are placed at points (-3 cm, 0, 0) and (3 cm, 0, 0) respectively. An external electric field \( \vec{E} = \frac{A}{r^2}\hat{r} \) where \( A = 3\times10^5 \) Vm is switched on in the region. Calculate the change in electrostatic energy of the system due to the electric field.
(ii) A system of two conductors is placed in air and they have net charge of +80μC and -80μC which causes a potential difference of 16 V between them.
(1) Find the capacitance of the system.
(2) If the air between the capacitor is replaced by a dielectric medium of dielectric constant 3, what will be the potential difference between the two conductors ?
(3) If the charges on two conductors are changed to +160 μC and -160 μC, will the capacitance of the system change ? Give reason for your answer. [5 Marks]

Answer:
1. Since \( \vec{E} = \frac{3\times10^5}{r^2}\hat{r} \) and \( dV = -\vec{E}\cdot d\vec{r} \), the potential is \( V = \frac{3\times10^5}{r} \).
2. Energy without the field \( U_i = \frac{Kq_1q_2}{r_{12}} \); with the field \( U_f = \frac{Kq_1q_2}{r_{12}} + q_1V(r_1) + q_2V(r_2) \). So \( \Delta U = U_f - U_i = q_1V(r_1) + q_2V(r_2) \).
3. \( \Delta U = \frac{5\times10^{-6}\times3\times10^5}{3\times10^{-2}} - \frac{1\times10^{-6}\times3\times10^5}{3\times10^{-2}} = 50 - 10 = 40 \) J.
4. (ii)(1) \( C = \frac{Q}{V} = \frac{80\ \mu C}{16\ V} = 5\ \mu F \). (2) \( C' = KC = 3\times5 = 15\ \mu F \), so \( V' = \frac{Q}{C'} = \frac{80\ \mu C}{15\ \mu F} = 5.33 \) V.
5. (ii)(3) No. The capacitance of the system depends only on its geometry (and the medium), not on the charge given to it.

Teacher's Note:
a) Both charges are at a distance of 3 cm from the origin, so \( r_1 = r_2 = 3\times10^{-2} \) m.
b) The mutual energy term cancels; only the energy of each charge in the external field changes.
c) Capacitance is a geometric property: doubling Q doubles V, and C stays the same.

OR

(b) (i) Consider three metal spherical shells A, B and C, each of radius R. Each shell is having a concentric metal ball of radius R/10. The spherical shells A, B and C are given charges +6q, -4q, and 14q respectively. Their inner metal balls are also given charges -2q, +8q and -10q respectively. Compare the magnitude of the electric fields due to shells A, B and C at a distance 3R from their centres.
(ii) A charge -6 μC is placed at the centre B of a semicircle of radius 5 cm, as shown in the figure. An equal and opposite charge is placed at point D at a distance of 10 cm from B. A charge +5 μC is moved from point 'C' to point 'A' along the circumference. Calculate the work done on the charge. [5 Marks]

[Figure: A straight line with points D, C, B and A from left to right. A semicircle of radius 5 cm with centre B is drawn above the line from C to A. Charge +6q is marked at D and -6q at B.]

Answer:
1. Total charge for A = 6q - 2q = +4q, for B = -4q + 8q = +4q, for C = 14q - 10q = +4q. So the total charge is the same (+4q) for all three.
2. At r = 3R, \( E = \frac{kQ}{r^2} = \frac{k(4q)}{9R^2} = \frac{4kq}{9R^2} \) for each, so \( E_A = E_B = E_C \).
3. At C (5 cm from both B and D): \( V_C = \frac{k\times6\times10^{-6}}{5\times10^{-2}} - \frac{k\times6\times10^{-6}}{5\times10^{-2}} = 0 \).
4. At A (15 cm from D, 5 cm from B): \( V_A = \frac{k\times6\times10^{-6}}{15\times10^{-2}} - \frac{k\times6\times10^{-6}}{5\times10^{-2}} = -\frac{9\times10^9\times6\times10^{-6}\times2}{15\times10^{-2}} = -7.2\times10^5 \) V.
5. Work done \( W = q(V_A - V_C) = 5\times10^{-6}\times(-7.2\times10^5 - 0) = -3.6 \) J.

Teacher's Note:
a) Outside a shell, the field depends only on the total charge enclosed (shell + inner ball).
b) Point C is equidistant from the equal and opposite charges, so its potential is zero.
c) Work done depends only on the end points, not on the semicircular path followed.

33. (a) (i) A proton moving with velocity \( \vec{V} \) in a non-uniform magnetic field traces a path as shown in the figure.
The path followed by the proton is always in the plane of the paper. What is the direction of the magnetic field in the region near points P, Q and R ? What can you say about relative magnitude of magnetic fields at these points ?
(ii) A current carrying circular loop of area A produces a magnetic field B at its centre. Show that the magnetic moment of the loop is \( \frac{2BA}{\mu_0}\sqrt{\frac{A}{\pi}} \). [5 Marks]

[Figure: A proton (+) with velocity \( \vec{V} \) moves horizontally to the right up to point P, then its path curves upward through point Q and reaches a crest at point R, after which it bends downward.]

Answer:
1. Near P and near Q, the force on the proton is upwards, so the magnetic field is into the plane of the paper.
2. Near R, the force \( \vec{F} \) is downwards, so the magnetic field is out of the plane of the paper.
3. Since \( B \propto \frac{1}{r} \) (r = radius of curvature of the path): near P, B is small; near Q, B is smaller than at P; near R, B is larger than at P. So \( B_Q \lt B_P \lt B_R \).
4. For the loop of radius r: \( B = \frac{\mu_0 I}{2r} \), so \( I = \frac{2Br}{\mu_0} \); and \( A = \pi r^2 \), so \( r = \sqrt{\frac{A}{\pi}} \).
5. \( M = IA = \frac{2Br}{\mu_0}A = \frac{2BA}{\mu_0}\sqrt{\frac{A}{\pi}} \).

Teacher's Note:
a) Use \( \vec{F} = q\vec{v}\times\vec{B} \) at each point; the force always points towards the centre of the curve.
b) The path is least curved near Q and most sharply curved near R, so B is weakest near Q and strongest near R.
c) Express r in terms of A before substituting in M = IA.

OR

(b) (i) Derive an expression for the torque acting on a rectangular current loop suspended in a uniform magnetic field.
(ii) A charged particle is moving in a circular path with velocity \( \vec{V} \) in a uniform magnetic field \( \vec{B} \). It is made to pass through a sheet of lead and as a consequence, it looses one half of its kinetic energy without change in its direction. How will (1) the radius of its path (2) its time period of revolution change ? [5 Marks]

Answer:
1. Consider a rectangular loop of sides a and b carrying current I, with its magnetic moment \( \vec{m} \) at angle \( \theta \) with \( \vec{B} \). The forces on the two arms of length b are \( |\vec{F_1}| = |\vec{F_2}| = IbB \), equal and opposite; the forces on the other two arms cancel.
2. F₁ and F₂ form a couple: \( \tau = F_1\frac{a}{2}\sin\theta + F_2\frac{a}{2}\sin\theta = IabB\sin\theta = IAB\sin\theta \).
3. In vector form, \( \vec{\tau} = I\vec{A}\times\vec{B} \) (for N turns, \( \tau = NIAB\sin\theta \)).
4. (ii)(1) \( r = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB} \), so \( r \propto \sqrt{K} \). \( \frac{r'}{r} = \frac{\sqrt{K/2}}{\sqrt{K}} = \frac{1}{\sqrt2} \), so the radius becomes \( r' = \frac{r}{\sqrt2} \).
5. (ii)(2) \( T = \frac{2\pi m}{qB} \) does not depend on kinetic energy, so the time period will not change.

Teacher's Note:
a) Draw a clear diagram showing F₁, F₂, \( \vec{B} \), \( \vec{m} \) and the perpendicular distance \( \frac{a}{2}\sin\theta \).
b) Halving K reduces the radius by a factor of \( \sqrt2 \), not by half.
c) The time period in a magnetic field is independent of speed and kinetic energy.

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